# A curious case of Relativistic Velocity Addition [duplicate]

This question already has an answer here:

The relativistic velocity addition formula is

$$u = \frac{v+u'}{1+ \frac{vu'}{c^2}}$$

Where

$u$ = velocity of projectile seen by rest observer "A"

$v$ = velocity of moving observer "B" as seen by rest observer "A"

$u'$ = velocity of projectile seen by B

Now the question is this: If $v=c,$ and $u'=-c$

I get an undefined answer. i.e. the relavistic velocity addition formula is undefined.

Whats wrong with setting $v=c,$ and $u'=-c$?

## marked as duplicate by Kyle Kanos, Ali, ACuriousMind♦, Brandon Enright, BernhardAug 26 '14 at 18:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• possible duplicate of Speed of light travel – Kyle Kanos Aug 26 '14 at 15:34
• An observer can not travel at speed $c$ with respect to anyone. – WillO Aug 26 '14 at 15:38

## 2 Answers

This isn't definitive, since the answer is always undefined, but let's be cutesy. Let's let $u' = a*c$ and $v = -a*c$, where $a < 1$

Then,

\begin{align} u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{ac-ac}{1- \frac{a^{2}c^{2}}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{c-c}{-2a}\\ &= 0 \end{align}

Now, the reason why this isn't definitive is that you can take different limits, if you want. Say, let $u' = a^{2}c$ and $v = -ac$

Then,

\begin{align} u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{a^{2}c-ac}{1+ \frac{a^{3}c^{2}}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{c\left(2a -1\right)}{3a^{2}}\\ &= \frac{c}{3} \end{align}

So, it's clear that, by taking the limit in different ways, you can get an arbitrary answer. It's not valid to choose an observer moving at the speed of light and then take velocities relative to that observer.

• @lurscher, your edit messed up the MathJax again! – Danu Aug 26 '14 at 15:51
• @lurscher, you broke my latex – Jerry Schirmer Aug 26 '14 at 15:51
• oh my. It was right before? It didn't render at all – lurscher Aug 26 '14 at 15:52
• one before last line: \frac{c}{3} (too short edit) – Void Aug 26 '14 at 15:57
• @lurscher: my first draft had broken latex, I fixed it, and then you edited. We must have been concurrently editing it. – Jerry Schirmer Aug 26 '14 at 16:01

It is important to think about what the velocity addition formula is meant to do. The velocity addition formula should be applied in situations of this type:

When you (person $A$) see someone (person $B$) speeding by you with velocity $v$, and when this second person sees another person (person $C$, one can also substitute objects for people, of course) pass with velocity $u'$, then what is the velocity $u$ at which person $A$ sees person $C$ come by?

As @WillO already pointed out in the comments, if person $B$ is speeding by at $c$, then there is no reference frame in which he/she/it is stationary. Therefore, it is clear that the velocity addition formula should not be applied here. Note that this only rules out situations where both $|v|$ and $|u'|$ equal $c$, since the formula is symmetrical under interchange of the two.

• If I set v = u' = c, I get u = c. It is a definite answer. However since you mentioned that person B cannot be speeding by at c, despite getting a definite u = c, can I also say that the calculation of setting v = u ' = c is wrong? – Candy Man Aug 26 '14 at 16:28
• I would say that it's just not meaningful to apply the velocity addition formula in that situation. – Danu Aug 26 '14 at 16:30
• Alright I see what you mean! – Candy Man Aug 26 '14 at 16:33